JEE MAIN - Physics (2024 - 29th January Morning Shift - No. 21)
In a double slit experiment shown in figure, when light of wavelength $$400 \mathrm{~nm}$$ is used, dark fringe is observed at $$P$$. If $$\mathrm{D}=0.2 \mathrm{~m}$$, the minimum distance between the slits $$S_1$$ and $$S_2$$ is _________ $$\mathrm{mm}$$.
_29th_January_Morning_Shift_en_21_1.png)
Explanation
Path difference for minima at $$\mathrm{P}$$
$$\begin{aligned} & 2 \sqrt{\mathrm{D}^2+\mathrm{d}^2}-2 \mathrm{D}=\frac{\lambda}{2} \\ & \therefore \sqrt{\mathrm{D}^2+\mathrm{d}^2}-\mathrm{D}=\frac{\lambda}{4} \\ & \therefore \sqrt{\mathrm{D}^2+\mathrm{d}^2}=\frac{\lambda}{4}+\mathrm{D} \\ & \Rightarrow \mathrm{D}^2+\mathrm{d}^2=\mathrm{D}^2+\frac{\lambda^2}{16}+\frac{\mathrm{D} \lambda}{2} \\ & \Rightarrow \mathrm{d}^2=\frac{\mathrm{D} \lambda}{2}+\frac{\lambda^2}{16} \\ & \Rightarrow \mathrm{d}^2=\frac{0.2 \times 400 \times 10^{-9}}{2}+\frac{4 \times 10^{-14}}{4} \\ & \Rightarrow \mathrm{d}^2 \approx 400 \times 10^{-10} \\ & \therefore \mathrm{d}=20 \times 10^{-5} \\ & \Rightarrow \mathrm{d}=0.20 \mathrm{~mm} \end{aligned}$$
Comments (0)
